Question 1(i)
In triangle ABC, right-angled at B, AB = 24 cm and BC = 7 cm. Find sin A and cos A.
Steps only
AC = √(24² + 7²) = 25
sin A = BC/AC = 7/25
cos A = AB/AC = 24/25
Answer
sin A = 7/25 and cos A = 24/25.
CLASS 10 · MATHEMATICS · CHAPTER 8
Introduction to Trigonometry
Build trigonometric ratios from right triangles, use exact standard-angle values and prove identities with concise or expanded working.
49 worked answersAll 3 exercise sets
What you will learn
- Identify all six trigonometric ratios.
- Use the standard values from 0° to 90°.
- Simplify and prove trigonometric identities.
49 worked answers · short steps first, explanations when you need them
Exercise 8.1
Textbook pages 121
Question 1(ii)
For the same triangle, find sin C and cos C.
Steps only
AC = 25
sin C = AB/AC = 24/25
cos C = BC/AC = 7/25
Answer
sin C = 24/25 and cos C = 7/25.
Question 2
In right triangle PQR, PQ = 12, QR = 5 and PR = 13. Find tan P − cot R.
Steps only
tan P = QR/PQ = 5/12
cot R = QR/PQ = 5/12
tan P − cot R = 0
Answer
0.
Question 3
If sin A = 3/4, calculate cos A and tan A.
Steps only
Opposite = 3k; hypotenuse = 4k
Adjacent = √(16k² − 9k²) = √7k
cos A = √7/4; tan A = 3/√7 = 3√7/7
Answer
cos A = √7/4 and tan A = 3√7/7.
Question 4
Given 15 cot A = 8, find sin A and sec A.
Steps only
cot A = 8/15, so adjacent:opposite = 8:15
Hypotenuse = √(8² + 15²) = 17
sin A = 15/17; sec A = 17/8
Answer
sin A = 15/17 and sec A = 17/8.
Question 5
Given sec θ = 13/12, calculate all the other trigonometric ratios.
Steps only
Hypotenuse:adjacent = 13:12, so opposite = √(13² − 12²) = 5
sin θ = 5/13; cos θ = 12/13; tan θ = 5/12
cosec θ = 13/5; cot θ = 12/5
Answer
sin θ = 5/13, cos θ = 12/13, tan θ = 5/12, cosec θ = 13/5 and cot θ = 12/5.
Question 6
If A and B are acute angles and cos A = cos B, show that A = B.
Steps only
Draw similar right triangles with acute angles A and B
Equal cosine ratios make their adjacent sides and hypotenuses proportional
Pythagoras makes the third sides proportional, so SSS similarity gives A = B
Answer
Hence, A = B.
Question 7(i)
If cot θ = 7/8, evaluate [(1 + sin θ)(1 − sin θ)]/[(1 + cos θ)(1 − cos θ)].
Steps only
Expression = (1 − sin²θ)/(1 − cos²θ)
= cos²θ/sin²θ = cot²θ
= (7/8)² = 49/64
Answer
49/64.
Question 7(ii)
If cot θ = 7/8, evaluate cot²θ.
Steps only
cot²θ = (7/8)²
= 49/64
Answer
49/64.
Question 8
If 3 cot A = 4, check whether (1 − tan²A)/(1 + tan²A) = cos²A − sin²A.
Steps only
cot A = 4/3, so tan A = 3/4, sin A = 3/5, cos A = 4/5
LHS = (1 − 9/16)/(1 + 9/16) = 7/25
RHS = 16/25 − 9/25 = 7/25
Answer
Yes. Both sides equal 7/25.
Question 9(i)
In triangle ABC, right-angled at B, tan A = 1/√3. Find sin A cos C + cos A sin C.
Steps only
A + C = 90°
sin A = 1/2, cos A = √3/2, cos C = 1/2, sin C = √3/2
Value = 1/4 + 3/4 = 1
Answer
1.
Question 9(ii)
For the same triangle, find cos A cos C − sin A sin C.
Steps only
cos A = √3/2; cos C = 1/2
sin A = 1/2; sin C = √3/2
Value = √3/4 − √3/4 = 0
Answer
0.
Question 10
In triangle PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Find sin P, cos P and tan P.
Steps only
Let QR = x, so PR = 25 − x
5² + x² = (25 − x)² ⇒ x = 12 and PR = 13
sin P = 12/13; cos P = 5/13; tan P = 12/5
Answer
sin P = 12/13, cos P = 5/13 and tan P = 12/5.
Question 11(i)
Is tan A always less than 1?
Steps only
tan 60° = √3
√3 > 1
Answer
False.
Question 11(ii)
Can sec A equal 12/5 for some acute angle A?
Steps only
sec A = hypotenuse/adjacent
A right triangle can have hypotenuse 12 and adjacent side 5
Answer
True.
Question 11(iii)
Is cos A the abbreviation for the cosecant of A?
Steps only
cos A means cosine of A
cosec A means cosecant of A
Answer
False.
Question 11(iv)
Is cot A the product of cot and A?
Steps only
cot A is one trigonometric ratio
The separated word “cot” is not a number
Answer
False.
Question 11(v)
Can sin θ equal 4/3 for an acute angle θ?
Steps only
sin θ = opposite/hypotenuse
opposite ≤ hypotenuse, so 0 < sin θ ≤ 1
4/3 > 1
Answer
False.
Exercise 8.2
Textbook pages 127
Question 1(i)
Evaluate sin 60° cos 30° + sin 30° cos 60°.
Steps only
= (√3/2)(√3/2) + (1/2)(1/2)
= 3/4 + 1/4 = 1
Answer
1.
Question 1(ii)
Evaluate 2 tan²45° + cos²30° − sin²60°.
Steps only
= 2(1)² + (√3/2)² − (√3/2)²
= 2
Answer
2.
Question 1(iii)
Evaluate cos 45°/(sec 30° + cosec 30°).
Steps only
= (1/√2)/(2/√3 + 2)
= (3√2 − √6)/8
Answer
(3√2 − √6)/8.
Question 1(iv)
Evaluate (sin 30° + tan 45° − cosec 60°)/(sec 30° + cos 60° + cot 45°).
Steps only
= (3/2 − 2/√3)/(3/2 + 2/√3)
= (3√3 − 4)/(3√3 + 4)
= (43 − 24√3)/11
Answer
(43 − 24√3)/11.
Question 1(v)
Evaluate [5 cos²60° + 4 sec²30° − tan²45°]/[sin²30° + cos²30°].
Steps only
Numerator = 5/4 + 16/3 − 1 = 67/12
Denominator = 1/4 + 3/4 = 1
Value = 67/12
Answer
67/12.
Question 2(i)
Choose the value of 2 tan 30°/(1 + tan²30°).
Steps only
= (2/√3)/(1 + 1/3)
= √3/2 = sin 60°
Answer
Option A: sin 60°.
Question 2(ii)
Choose the value of (1 − tan²45°)/(1 + tan²45°).
Steps only
= (1 − 1)/(1 + 1)
= 0
Answer
Option D: 0.
Question 2(iii)
For which listed angle is sin 2A = 2 sin A true?
Steps only
At A = 0°: sin 0° = 0 and 2 sin 0° = 0
Answer
Option A: 0°.
Question 2(iv)
Choose the value of 2 tan 30°/(1 − tan²30°).
Steps only
= (2/√3)/(1 − 1/3)
= √3 = tan 60°
Answer
Option C: tan 60°.
Question 3
If tan(A + B) = √3 and tan(A − B) = 1/√3, with 0° < A + B ≤ 90° and A > B, find A and B.
Steps only
A + B = 60°
A − B = 30°
2A = 90° ⇒ A = 45°; B = 15°
Answer
A = 45° and B = 15°.
Question 4(i)
Is sin(A + B) = sin A + sin B?
Steps only
Take A = B = 30°
sin 60° = √3/2, while sin 30° + sin 30° = 1
Answer
False.
Question 4(ii)
Does sin θ increase as θ increases from 0° to 90°?
Steps only
sin 0° = 0 < sin 30° = 1/2 < sin 45° = 1/√2 < sin 60° = √3/2 < sin 90° = 1
Answer
True.
Question 4(iii)
Does cos θ increase as θ increases from 0° to 90°?
Steps only
cos 0° = 1
cos 90° = 0
Answer
False; cosine decreases on this interval.
Question 4(iv)
Is sin θ = cos θ for every θ?
Steps only
At θ = 30°, sin θ = 1/2
At θ = 30°, cos θ = √3/2
Answer
False; they are equal only at 45° in this interval.
Question 4(v)
Is cot A undefined when A = 0°?
Steps only
cot 0° = cos 0°/sin 0°
= 1/0, which is undefined
Answer
True.
Exercise 8.3
Textbook pages 131–132
Question 1
Express sin A, sec A and tan A in terms of cot A.
Steps only
cosec²A = 1 + cot²A
sin A = 1/√(1 + cot²A)
sec A = √(1 + cot²A)/cot A; tan A = 1/cot A
Answer
sin A = 1/√(1 + cot²A), sec A = √(1 + cot²A)/cot A and tan A = 1/cot A.
Question 2
Write all other trigonometric ratios of A in terms of sec A.
Steps only
cos A = 1/sec A; tan A = √(sec²A − 1)
sin A = √(sec²A − 1)/sec A
cot A = 1/√(sec²A − 1); cosec A = sec A/√(sec²A − 1)
Answer
cos A = 1/sec A, sin A = √(sec²A − 1)/sec A, tan A = √(sec²A − 1), cot A = 1/√(sec²A − 1), cosec A = sec A/√(sec²A − 1).
Question 3(i)
Choose the value of 9 sec²A − 9 tan²A.
Steps only
= 9(sec²A − tan²A)
= 9(1) = 9
Answer
Option B: 9.
Question 3(ii)
Choose the value of (1 + tan θ + sec θ)(1 + cot θ − cosec θ).
Steps only
= [(sin θ + cos θ + 1)/cos θ][(sin θ + cos θ − 1)/sin θ]
= [(sin θ + cos θ)² − 1]/(sin θ cos θ)
= 2 sin θ cos θ/(sin θ cos θ) = 2
Answer
Option C: 2.
Question 3(iii)
Choose the value of (sec A + tan A)(1 − sin A).
Steps only
= [(1 + sin A)/cos A](1 − sin A)
= (1 − sin²A)/cos A
= cos A
Answer
Option D: cos A.
Question 3(iv)
Choose the value of (1 + tan²A)/(1 + cot²A).
Steps only
= sec²A/cosec²A
= sin²A/cos²A
= tan²A
Answer
Option D: tan²A.
Question 4(i)
Prove (cosec θ − cot θ)² = (1 − cos θ)/(1 + cos θ).
Steps only
LHS = [(1 − cos θ)/sin θ]²
= (1 − cos θ)²/[(1 − cos θ)(1 + cos θ)]
= (1 − cos θ)/(1 + cos θ) = RHS
Answer
Identity proved.
Question 4(ii)
Prove cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A.
Steps only
LHS = [cos²A + (1 + sin A)²]/[(1 + sin A)cos A]
= [2 + 2 sin A]/[(1 + sin A)cos A]
= 2/cos A = 2 sec A = RHS
Answer
Identity proved.
Question 4(iii)
Prove tan θ/(1 − cot θ) + cot θ/(1 − tan θ) = 1 + sec θ cosec θ.
Steps only
LHS = [sin³θ − cos³θ]/[sin θ cos θ(sin θ − cos θ)]
= [sin²θ + sin θ cos θ + cos²θ]/(sin θ cos θ)
= 1 + 1/(sin θ cos θ) = 1 + sec θ cosec θ
Answer
Identity proved.
Question 4(iv)
Prove (1 + sec A)/sec A = sin²A/(1 − cos A).
Steps only
LHS = cos A + 1
RHS = (1 − cos²A)/(1 − cos A)
= 1 + cos A = LHS
Answer
Identity proved.
Question 4(v)
Prove (cos A − sin A + 1)/(cos A + sin A − 1) = cosec A + cot A.
Steps only
Divide numerator and denominator by sin A
LHS = (cot A − 1 + cosec A)/(cot A + 1 − cosec A)
Use 1 = (cosec A − cot A)(cosec A + cot A); cancellation gives cosec A + cot A
Answer
Identity proved.
Question 4(vi)
Prove √[(1 + sin A)/(1 − sin A)] = sec A + tan A.
Steps only
LHS = √[(1 + sin A)²/(1 − sin²A)]
= (1 + sin A)/cos A
= sec A + tan A = RHS
Answer
Identity proved.
Question 4(vii)
Prove (sin θ − 2 sin³θ)/(2 cos³θ − cos θ) = tan θ.
Steps only
LHS = sin θ(1 − 2 sin²θ)/[cos θ(2 cos²θ − 1)]
1 − 2 sin²θ = 2 cos²θ − 1
LHS = sin θ/cos θ = tan θ
Answer
Identity proved.
Question 4(viii)
Prove (sin A + cosec A)² + (cos A + sec A)² = 7 + tan²A + cot²A.
Steps only
LHS = sin²A + cos²A + 4 + cosec²A + sec²A
= 1 + 4 + (1 + cot²A) + (1 + tan²A)
= 7 + tan²A + cot²A
Answer
Identity proved.
Question 4(ix)
Prove (cosec A − sin A)(sec A − cos A) = 1/(tan A + cot A).
Steps only
LHS = (cos²A/sin A)(sin²A/cos A) = sin A cos A
RHS = 1/[sin A/cos A + cos A/sin A]
= sin A cos A/(sin²A + cos²A) = sin A cos A
Answer
Identity proved.
Question 4(x)
Prove (1 + tan²A)/(1 + cot²A) = [(1 − tan A)/(1 − cot A)]² = tan²A.
Steps only
First expression = sec²A/cosec²A = tan²A
(1 − tan A)/(1 − cot A) = (1 − tan A)/(1 − 1/tan A) = −tan A
Squaring gives tan²A
Answer
All three expressions are equal to tan²A.